Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $x = \dfrac{k^2 - 8k + 16}{k - 4} \div \dfrac{-7k + 28}{k - 6} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{k^2 - 8k + 16}{k - 4} \times \dfrac{k - 6}{-7k + 28} $ First factor the quadratic. $x = \dfrac{(k - 4)(k - 4)}{k - 4} \times \dfrac{k - 6}{-7k + 28} $ Then factor out any other terms. $x = \dfrac{(k - 4)(k - 4)}{k - 4} \times \dfrac{k - 6}{-7(k - 4)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (k - 4)(k - 4) \times (k - 6) } { (k - 4) \times -7(k - 4) } $ $x = \dfrac{ (k - 4)(k - 4)(k - 6)}{ -7(k - 4)(k - 4)} $ Notice that $(k - 4)$ appears twice in both the numerator and denominator so we can cancel them. $x = \dfrac{ \cancel{(k - 4)}(k - 4)(k - 6)}{ -7(k - 4)\cancel{(k - 4)}} $ We are dividing by $k - 4$ , so $k - 4 \neq 0$ Therefore, $k \neq 4$ $x = \dfrac{ \cancel{(k - 4)}\cancel{(k - 4)}(k - 6)}{ -7\cancel{(k - 4)}\cancel{(k - 4)}} $ We are dividing by $k - 4$ , so $k - 4 \neq 0$ Therefore, $k \neq 4$ $x = \dfrac{k - 6}{-7} $ $x = \dfrac{-(k - 6)}{7} ; \space k \neq 4 $